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Mathematics Question on Sum of First n Terms of an AP

Sum of n terms of the series 1+11+111+... 1 + 11 + 111 + ... is

A

1081(10n−1)−n81\frac{10}{81}(10^n-1)-\frac{n}{81}

B

1081(10n−1)−n9\frac{10}{81}(10^n-1)-\frac{n}{9}

C

1081(10n−1)+n9\frac{10}{81}(10^n-1)+\frac{n}{9}

D

1081(10n−1)+n81\frac{10}{81}(10^n-1)+\frac{n}{81}

Answer

1081(10n−1)−n9\frac{10}{81}(10^n-1)-\frac{n}{9}

Explanation

Solution

Sn=1+11+111+.....nS_{n} = 1+11+111+.....n terms =19[9+99+999...n= \frac{1}{9}[9+99+999...n terms] =19[(10−1)+(102−1)+(103−1)+...= \frac{1}{9} [\left(10-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+...) =19[10(10n−1)10−1−n] = \frac{1}{9}\left[10 \frac{\left(10^{n}-1\right)}{10-1}-n\right] =1081[10n−1]−n9= \frac{10}{81}\left[10^{n}-1\right]-\frac{n}{9}